This was a revised question from the post and based on an example provided by user779041.
It's been wildly known the mathematics of Dirac delta function, which was thought not a function, but a distribution, that all the "probability distribution" went to a "point".(or as a generalization of the Gaussian distribution)
$$p(x=0)=\lim_{n\rightarrow 0} \frac{1}{n}$$
However, consider the counterpart of such procedure, that the distribution was to spread out over the entire real axis, with infinite variance.
$$p(x|x\in(-\frac{1}{n},\frac{1}{n}))=\lim_{n\rightarrow \infty} \frac{1}{n}$$
Compare with the Cauchy distribution, which also had infinite variance, one make further requirement that, at any finite measure of interval, the measure over the distribution was zero, i.e.
$$\int_{x\in[a,b]} dx p(x)=0$$
for some fixed $a,b\in\mathbb{R}$.The procedure through limiting $\frac{1}{n}$ could make some sense, or equivalently taking the limit of $\sigma \rightarrow \infty$ in the Gaussian distribution.
However, unlike Dirac delta function, the new "reversed function"(say $\theta(x)$) was over an interval. With interval, there should be able to assign "shape"(say a "localized" function $s(x)$).
But how to assign such shape function to the $\theta(x)$, so that $\theta(x)$ could have "shape"/"weights" over the interval? How does the multiplication of two of such shaped $\theta(x)$ work?
*Extra point but not quite sure: the finite discontinuity in a shape function $s(x)$ might be ignored or smoothed out with the $\theta(x)$ if it works.